Math 2250 Written HW #10 Solutions
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1. Find the absolute maximum and minimum values of the function g(x) = e−x subject to the
constraint −2 ≤ x ≤ 1.
Answer: First, we check for critical points of g, which is differentiable everywhere. By the
Chain Rule,
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g 0 (x) = e−x · (−2x) = −2xe−x .
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Since e−x > 0 for all x, g 0 (x) = 0 when 2x = 0, meaning when x = 0. Hence, x = 0 is the
only critical point of g. Now we just evaluate g at the critical point and the endpoints:
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g(−2) = e−(−2) = e−4 = 1/e4
2
g(0) = e−0 = e0 = 1
2
g(2) = e−1 = e−1 = 1/e.
Since 1 > 1/e > 1/e4 , we see that the absolute maximum of g(x) on this interval is at (0, 1)
and the absolute minimum is at (−2, 1/e4 ).
2. Find all local maxima and minima of the curve y = x2 ln x.
Answer: Notice, first of all, that ln x is only defined for x > 0, so the function x2 ln x is
likewise only defined for x > 0.
This function is differentiable on its entire domain, so we differentiate in search of critical
points. Using the Product Rule,
y 0 = 2x ln x + x2 ·
1
= 2x ln x + x = x(2 ln x + 1).
x
Since we’re only allowed to consider x > 0, we see that the derivative is zero only when
2 ln x + 1 = 0, meaning ln x = −1/2. Therefore, we have a critical point at x = e−1/2 .
For x > 0, the expression x(2 ln x + 1) has the same sign as 2 ln x + 1, which is negative for
x < e−1/2 and positive otherwise. Therefore, the curve has a local minimum when x = e−1/2 ,
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meaning at the point ( √1e , − 2e
). There are no other local extrema of this curve.
3. A general cubic function has the form
f (x) = ax3 + bx2 + cx + d
where a, b, c, and d are constants.
(a) Give examples that demonstrate such functions can have 0, 1, or 2 critical points.
Answer: Suppose f (x) = x3 + x. Then f 0 (x) = 3x2 + 1, which is always positive, so
this is an example of a cubic function with no critical points.
If f (x) = x3 , then f 0 (x) = 3x2 , which is zero only when x = 0, so this is a cubic function
with exactly 1 critical point.
Finally, if f (x) = x3 + x2 − x + 1, then f 0 (x) = 3x2 + 2x − 1 = (3x − 1)(x + 1), so this
cubic function has critical points when x = 1/3 and x = −1. Thus, this cubic function
has exactly two critical points.
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(b) Show that no cubic function can have more than 2 critical points.
Answer: For any cubic function f (x) = ax3 + bx2 + cx + d, we know that
f 0 (x) = 3ax2 + 2bx + c.
Therefore, f 0 (x) = 0 if and only if
p
√
−2b ± (2b)2 − 4(3a)(c)
−2b ± 4b2 − 12ac
=
.
x=
2(3a)
6a
There are at most such two values of x, so we see that f has at most two critical points.
(c) How many local extreme values (maxima and/or minima) can a cubic function have?
Answer: A cubic function can certainly have no local extreme values, as in the case of
f (x) = x3 + x, which we saw in part (a) has no critical points.
A cubic function can also have two local extreme values (1 max and 1 min), as in the
case of f (x) = x3 + x2 + x + 1, which has a local maximum at x = −1 and a local
minimum at x = 1/3.
These are the only options. Since a cubic function can’t have more than two critical
points, it certainly can’t have more than two extreme values. Also, a cubic function
cannot have just one local extremum except in the slightly dumb case when a = 0 (in
which case it’s really a quadratic function instead of a cubic).
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